1) A compound gives a mass spectrum with peaks at...

1) A compound gives a mass spectrum with peaks at m/z = 77
(40%), 112 (100%), 114 (33%), and essentially no other peaks.
Identify the compound.
First, your molecular ion peak is 112 and you have a M+2 peak at
114. Therefore, you have a halogen.
Now, your molecular ion peak and M+2 peak are in a 3 to 1 ratio.
This means chlorine.
So, 112-35=77
# C’s 77/12=6 carbons so C6H5Cl. DOUS (2(6)+2-5-1)/2=4
Cl
2) While organizing the undergraduate stockroom, a new chemistry
professor found a half-gallon jug containing a cloudy liquid (bp
100–105 °C), marked only "STUDENT PREP". She ran a quick mass
spectrum, which is shown below. As soon as she saw the spectrum
(without even checking the actual mass numbers), she said, "I know
what it is." What compound is the "student prep"?
So molecular ion peak at 136 and M+2 peak at
138, so halogen present.
They are in a 1:1 ratio so Br.
So 136-79=57/12 = 4x12=48
57-48=9, C4H9Br (2(4)+2-9-1)/2=0
Br
The peaks at 107 (C2H5) and 93(C3H7)
tell us it is a linear chain instead of a
branched one.
3) A laboratory student added 1-bromobutane to a flask
containing dry ether and Mg turnings. An exothermic reaction
resulted, and the ether boiled vigorously for several minutes. Then
she added acetone to the reaction mixture, and the ether boiled
even more vigorously. She added dilute acid to the mixture, and
separated the layers. She evaporated the ether layer, and distilled a
liquid that boiled at 143 °C. GC–MS analysis of the distillate
showed one major product with a few minor impurities. The mass
spectrum of the major product is shown below. Show the structure
of this major product.
BrMg/etherMgBrOO-H+OH
The molecular ion peak should be at 116, but the loss of a
carbon from the quatenary C forms a stable carbocation at
101.
1) One of the following compounds is responsible for the IR
spectrum shown. Draw the structure of the responsible
compound.
1-butene, 1-butanol, 4-hydroxy-1-butene, methyl propyl ether,
butanoic acid.
First thing to notice is the presence of an alcohol at 3200-3600 cm-1,
which narrows our choices down to 1-butanol, 4-hydroxy-1-butene
and butanoic acid..
OHO
OH
OH
Only 1-butanol works, because there are no peaks
corresponding to C=C and C=O.
2) One of the following compounds is responsible for the IR
spectrum shown. Draw the structure of the responsible compound.
phenylacetone, benzoic acid, acetophenone, benzyl alcohol,
benzaldehyde
First you can eliminate benzoic acid and benzyl alcohol because
there is no –OH peak.
Second you can eliminate benzaldehyde because there is no peak
at 2740 cm-1 (aldehyde peak).
That leaves phenylacetone and acetophenone.
OO
Acetophenone is the answer because the carbonyl peak is at
1700 cm-1 and a simple ketone like that on phenylacetone
would absorb at a higher energy.
3) One of the following compounds is responsible for the IR
spectrum shown. Draw the structure of the responsible compound.
2-ethynylcyclohexanone, 2-methyl-2-cyclohexenone,
acetophenone, cyclohexylmethyl alcohol, 4-ethylcyclohexanone.
First, 2-ethynylcyclohexanone can be eliminated because there is
no peak for a carbon carbon triple bond.
Second, cyclohexylmethyl alcohol is eliminated because there is
no –OH peak present.
Acetopheone can be eliminated next because there is no peak for
aromatic C-H stretches. Also the carbonyl peak would be at a
higher energy like around 1800 cm-1.
2-methyl-2-cyclohexenone can be eliminatedObecause there is no
carbon carbon double bond peak.
Leaving 4-ethylcyclohexanone.
(4) Determine the structure of the compound that
gives rise to the following mass and IR spectra.
The molecular ion peak is at 162 and the M+2 peak is at 164, and
they are in a 1:1 ratio, therefore there is a bromine atom.
162-79=83
83/12=6 carbons and 6x12= 72 so 83-72=11 C6H11Br
So 2(6)+2-11-1=2/2=1, which means 1 double bond or 1 ring.
Looking at the IR, there is no C=C peak so that means a ring.
Br
(5) Determine the structure of the compound that gives rise to the
following mass and IR spectra.
So the molecular ion peak is 72.
So 72/12= 6 carbons 2(6)+2=14/2=7 a bit high.
So subtract 1 C and replace with 12 Hs. C5H12.
2(5)+2-12=0
Not pentane, there is a carbonyl stretch.
So to add an O, subtract a methyl group C4H8O.
2(4)+2-8=2/2=1
Lastly there is a peak at 2740 which tells us that the carbonyl si due
to an aldehyde.
O
(6) Draw a structure consistent with the following data:
•The MS shows a molecular ion at 59 amu.
•The IR spectrum shows a double-humped strong absorbance at
around 3300 cm–1 (the only absorbance in the functional group
region) and a single absorbance at about 1385 cm–1.
Odd molecular ion peak tells you there is a nitrogen.
59-14=45 45/12=3 carbons 45-36=9 hydrogens
C3H9N. 2(3)+2-9+1=0
The peak at 3300 cm–1 tells us that the N is part of an amine.
NH2
(1) The following 1H NMR spectrum is of a compound of molecular
formula C3H8O. Propose a structure for this compound.
First you have a septet that integrates to 1 H and a doublet that
integrates to 6 Hs. This is typical of an isopropyl group.
Then the peak at 2.5ppm. Is a singlet and represents a H on an OH
group.
OH
2) Draw the structure of the compound with the 1H NMR and IR
spectra shown and the formula C5H12O.
2(5)+2-12=0 so no double bonds or rings.
Also there is no –OH or C=O peaks in the IR, so it has to be an
ether.
Looking downfield you have a triplet and a singlet. For there to
be a singlet there must be only a methyl on one side of the ether.
Thus giving us the following structure.
O
Looking at this structure, it explains the presence of the pentet
and sextet for the middle two CH2s. And finally the triplet
upfield is for the terminal methyl group.
3) Draw the structure of the compound with the 1H NMR and IR
spectra shown and the formula C6H12O2.
First 2(6)+2-12=2/2=1
And based on the carbonyl peak in the IR we know this is our
degree of unsaturation.
Also we know that there must also be an ether since there is no
OH peak in the IR.
And based on the proton NMR we have two types of protons.
One has to be connected to the ether.
And for the rest to be all the same, there must be an isobutyl
group.
OO
(4) Determine the structure of the following compound based on its
mass, IR, and 1H NMR spectra.
114/12=9 carbons
114-108=6 hydrogens
C9H6 2(9)+2-6= 7 degrees of unsaturation
Based on the IR we know there is a carbonyl
So –CH4 add O C8H2O 2(8)+2-2=16/2=8
Lets take off a C and add 12 Hs C7H14O 2(7)+2-14=2/2=1
Which accounts for the C=O
And now look at the proton NMR and there are 3 types of
protons, this indicates a symmetrical ketone. There should be
two triplets and one sextet.
O
(5) Either 2-butanone, 2-methyl-2-nitropropane, 3-pentanone, 1nitropropane, nitroethane, or 2-bromopropane is responsible for the 1H
NMR spectrum shown. Draw the structure of the responsible
compound.
There are two signals in the spectra so we can eliminate 2butanone, 2-methyl-2-nitropropane and 1-nitropropane because
they have either more or less than 2 types of protons.
Next you can eliminate 3-pentanone because it would have a
triplet and a quartet which is not seen in the spectra.
Also nitroethane can be eliminated because it would have a
doublet and a quartet.
While the answer is 2-bromopropane which has a doublet and a
heptet.
Br
6) Either 2-butanone, 2-methyl-2-nitropropane, 3-pentanone, 1nitropropane, nitroethane, or 2-bromopropane is responsible for
the 1H NMR spectrum shown. Draw the structure of the
responsible compound.
Because there are 3 signals it is either 2-butanone or 1nitropropane.
1-nitropropane should have a doublet, a quartet and a triplet.
While our answer, 2-butanone should have a singlet, a quartet
and a triplet..
O
7) The molecular formula of a compound is C6H12O. Determine the
structure of the compound based on its molecular formula and its
13C NMR spectrum.
First 2(6)+2-12=2/2=1 so either a ring or double bond.
No peak shows up in the double bond region, C=C or C=O.
So that leaves a ring.
Four peaks and with this structure we have 4 different types of
carbon.
OH
8) Identify the compound with molecular formula C3H5Cl3 that
gives the following 13C NMR spectrum. (The resonance at 0 ppm is
due to the TMS standard, not the unknown.)
First, 2(3)+2-5-3=0 so no double bonds or rings.
Secondly there are 3 peaks, so 3 different kinds fo carbon.
So that leaves two choices that are correct.
ClClClClClCl